Methods for finding the inverse matrix, . Consider a square matrix

Let us denote Δ =det A.

The square matrix A is called non-degenerate, or not special, if its determinant is nonzero, and degenerate, or special, IfΔ = 0.

A square matrix B is for a square matrix A of the same order if their product is A B = B A = E, where E is the identity matrix of the same order as the matrices A and B.

Theorem . In order for matrix A to have an inverse matrix, it is necessary and sufficient that its determinant be different from zero.

The inverse matrix of matrix A, denoted by A- 1, so B = A - 1 and is calculated by the formula

, (1)

where A i j are algebraic complements of elements a i j of matrix A..

Calculating A -1 using formula (1) for high-order matrices is very labor-intensive, so in practice it is convenient to find A -1 using the method of elementary transformations (ET). Any non-singular matrix A can be reduced to the identity matrix E by means of EDs of only columns (or only rows). If the EDs perfected over matrix A are applied in the same order to the identity matrix E, then the result is an inverse matrix. It is convenient to perform EP on matrices A and E simultaneously, writing both matrices side by side through a line. Let us note once again that when searching for the canonical form of a matrix, in order to find it, you can use transformations of rows and columns. If you need to find the inverse of a matrix, you should use only rows or only columns during the transformation process.

Example 2.10. For matrix find A -1 .

Solution.First we find the determinant of matrix A
This means that the inverse matrix exists and we can find it using the formula: , where A i j (i,j=1,2,3) are algebraic additions of elements a i j of the original matrix.

Where .

Example 2.11. Using the method of elementary transformations, find A -1 for the matrix: A = .

Solution.We assign to the original matrix on the right an identity matrix of the same order: . Using elementary transformations of the columns, we will reduce the left “half” to the identity one, simultaneously performing exactly the same transformations on the right matrix.
To do this, swap the first and second columns: ~ . To the third column we add the first, and to the second - the first, multiplied by -2: . From the first column we subtract the second doubled, and from the third - the second multiplied by 6; . Let's add the third column to the first and second: . Multiply the last column by -1: . The square matrix obtained to the right of the vertical bar is the inverse matrix of the given matrix A. So,
.

The matrix $A^(-1)$ is called the inverse of the square matrix $A$ if the condition $A^(-1)\cdot A=A\cdot A^(-1)=E$ is satisfied, where $E $ is the identity matrix, the order of which is equal to the order of the matrix $A$.

A non-singular matrix is ​​a matrix whose determinant is not equal to zero. Accordingly, a singular matrix is ​​one whose determinant is equal to zero.

The inverse matrix $A^(-1)$ exists if and only if the matrix $A$ is non-singular. If the inverse matrix $A^(-1)$ exists, then it is unique.

There are several ways to find the inverse of a matrix, and we will look at two of them. This page will discuss the adjoint matrix method, which is considered standard in most higher mathematics courses. The second method of finding the inverse matrix (the method of elementary transformations), which involves using the Gauss method or the Gauss-Jordan method, is discussed in the second part.

Adjoint matrix method

Let the matrix $A_(n\times n)$ be given. In order to find the inverse matrix $A^(-1)$, three steps are required:

1. Find the determinant of the matrix $A$ and make sure that $\Delta A\neq 0$, i.e. that matrix A is non-singular.
2. Compose algebraic complements $A_(ij)$ of each element of the matrix $A$ and write the matrix $A_(n\times n)^(*)=\left(A_(ij) \right)$ from the found algebraic complements.
3. Write the inverse matrix taking into account the formula $A^(-1)=\frac(1)(\Delta A)\cdot (A^(*))^T$.

The matrix $(A^(*))^T$ is often called adjoint (reciprocal, allied) to the matrix $A$.

If the solution is done manually, then the first method is good only for matrices of relatively small orders: second (), third (), fourth (). To find the inverse of a higher order matrix, other methods are used. For example, the Gaussian method, which is discussed in the second part.

Example No. 1

Find the inverse of matrix $A=\left(\begin(array) (cccc) 5 & -4 &1 & 0 \\ 12 &-11 &4 & 0 \\ -5 & 58 &4 & 0 \\ 3 & - 1 & -9 & 0 \end(array) \right)$.

Since all elements of the fourth column are equal to zero, then $\Delta A=0$ (i.e. the matrix $A$ is singular). Since $\Delta A=0$, there is no inverse matrix to matrix $A$.

Example No. 2

Find the inverse of matrix $A=\left(\begin(array) (cc) -5 & 7 \\ 9 & 8 \end(array)\right)$.

We use the adjoint matrix method. First, let's find the determinant of the given matrix $A$:

$$ \Delta A=\left| \begin(array) (cc) -5 & 7\\ 9 & 8 \end(array)\right|=-5\cdot 8-7\cdot 9=-103. $$

Since $\Delta A \neq 0$, then the inverse matrix exists, therefore we will continue the solution. Finding algebraic complements

\begin(aligned) & A_(11)=(-1)^2\cdot 8=8; \; A_(12)=(-1)^3\cdot 9=-9;\\ & A_(21)=(-1)^3\cdot 7=-7; \; A_(22)=(-1)^4\cdot (-5)=-5.\\ \end(aligned)

We compose a matrix of algebraic additions: $A^(*)=\left(\begin(array) (cc) 8 & -9\\ -7 & -5 \end(array)\right)$.

We transpose the resulting matrix: $(A^(*))^T=\left(\begin(array) (cc) 8 & -7\\ -9 & -5 \end(array)\right)$ (the resulting matrix is ​​often is called the adjoint or allied matrix to the matrix $A$). Using the formula $A^(-1)=\frac(1)(\Delta A)\cdot (A^(*))^T$, we have:

$$ A^(-1)=\frac(1)(-103)\cdot \left(\begin(array) (cc) 8 & -7\\ -9 & -5 \end(array)\right) =\left(\begin(array) (cc) -8/103 & 7/103\\ 9/103 & 5/103 \end(array)\right) $$

So, the inverse matrix is ​​found: $A^(-1)=\left(\begin(array) (cc) -8/103 & 7/103\\ 9/103 & 5/103 \end(array)\right) $. To check the truth of the result, it is enough to check the truth of one of the equalities: $A^(-1)\cdot A=E$ or $A\cdot A^(-1)=E$. Let's check the equality $A^(-1)\cdot A=E$. In order to work less with fractions, we will substitute the matrix $A^(-1)$ not in the form $\left(\begin(array) (cc) -8/103 & 7/103\\ 9/103 & 5/103 \ end(array)\right)$, and in the form $-\frac(1)(103)\cdot \left(\begin(array) (cc) 8 & -7\\ -9 & -5 \end(array )\right)$:

Answer: $A^(-1)=\left(\begin(array) (cc) -8/103 & 7/103\\ 9/103 & 5/103 \end(array)\right)$.

Example No. 3

Find the inverse matrix for the matrix $A=\left(\begin(array) (ccc) 1 & 7 & 3 \\ -4 & 9 & 4 \\ 0 & 3 & 2\end(array) \right)$.

Let's start by calculating the determinant of the matrix $A$. So, the determinant of the matrix $A$ is:

$$ \Delta A=\left| \begin(array) (ccc) 1 & 7 & 3 \\ -4 & 9 & 4 \\ 0 & 3 & 2\end(array) \right| = 18-36+56-12=26. $$

Since $\Delta A\neq 0$, then the inverse matrix exists, therefore we will continue the solution. We find the algebraic complements of each element of a given matrix:

We compose a matrix of algebraic additions and transpose it:

$$ A^*=\left(\begin(array) (ccc) 6 & 8 & -12 \\ -5 & 2 & -3 \\ 1 & -16 & 37\end(array) \right); \; (A^*)^T=\left(\begin(array) (ccc) 6 & -5 & 1 \\ 8 & 2 & -16 \\ -12 & -3 & 37\end(array) \right) $$

Using the formula $A^(-1)=\frac(1)(\Delta A)\cdot (A^(*))^T$, we get:

$$ A^(-1)=\frac(1)(26)\cdot \left(\begin(array) (ccc) 6 & -5 & 1 \\ 8 & 2 & -16 \\ -12 & - 3 & 37\end(array) \right)= \left(\begin(array) (ccc) 3/13 & -5/26 & 1/26 \\ 4/13 & 1/13 & -8/13 \ \ -6/13 & -3/26 & 37/26 \end(array) \right) $$

So $A^(-1)=\left(\begin(array) (ccc) 3/13 & -5/26 & 1/26 \\ 4/13 & 1/13 & -8/13 \\ - 6/13 & -3/26 & 37/26 \end(array) \right)$. To check the truth of the result, it is enough to check the truth of one of the equalities: $A^(-1)\cdot A=E$ or $A\cdot A^(-1)=E$. Let's check the equality $A\cdot A^(-1)=E$. In order to work less with fractions, we will substitute the matrix $A^(-1)$ not in the form $\left(\begin(array) (ccc) 3/13 & -5/26 & 1/26 \\ 4/13 & 1/13 & -8/13 \\ -6/13 & -3/26 & 37/26 \end(array) \right)$, and in the form $\frac(1)(26)\cdot \left( \begin(array) (ccc) 6 & -5 & 1 \\ 8 & 2 & -16 \\ -12 & -3 & 37\end(array) \right)$:

The check was successful, the inverse matrix $A^(-1)$ was found correctly.

Answer: $A^(-1)=\left(\begin(array) (ccc) 3/13 & -5/26 & 1/26 \\ 4/13 & 1/13 & -8/13 \\ -6 /13 & -3/26 & 37/26 \end(array) \right)$.

Example No. 4

Find the matrix inverse of matrix $A=\left(\begin(array) (cccc) 6 & -5 & 8 & 4\\ 9 & 7 & 5 & 2 \\ 7 & 5 & 3 & 7\\ -4 & 8 & -8 & -3 \end(array) \right)$.

For a fourth-order matrix, finding the inverse matrix using algebraic additions is somewhat difficult. However, such examples do occur in test papers.

To find the inverse of a matrix, you first need to calculate the determinant of the matrix $A$. The best way to do this in this situation is by decomposing the determinant along a row (column). We select any row or column and find the algebraic complements of each element of the selected row or column.

Definition 1: a matrix is ​​called singular if its determinant is zero.

Definition 2: a matrix is ​​called non-singular if its determinant is not equal to zero.

Matrix "A" is called inverse matrix, if the condition A*A-1 = A-1 *A = E (unit matrix) is satisfied.

A square matrix is ​​invertible only if it is non-singular.

Scheme for calculating the inverse matrix:

1) Calculate the determinant of matrix "A" if ∆ A = 0, then the inverse matrix does not exist.

2) Find all algebraic complements of matrix "A".

3) Create a matrix of algebraic additions (Aij)

4) Transpose the matrix of algebraic complements (Aij )T

5) Multiply the transposed matrix by the inverse of the determinant of this matrix.

6) Perform check:

At first glance it may seem complicated, but in fact everything is very simple. All solutions are based on simple arithmetic operations, the main thing when solving is not to get confused with the “-” and “+” signs and not to lose them.

Now let’s solve a practical task together by calculating the inverse matrix.

Task: find the inverse matrix "A" shown in the picture below:

We solve everything exactly as indicated in the plan for calculating the inverse matrix.

1. The first thing to do is to find the determinant of matrix "A":

Explanation:

We have simplified our determinant using its basic functions. First, we added to the 2nd and 3rd lines the elements of the first line, multiplied by one number.

Secondly, we changed the 2nd and 3rd columns of the determinant, and according to its properties, we changed the sign in front of it.

Thirdly, we took out the common factor (-1) of the second line, thereby changing the sign again, and it became positive. We also simplified line 3 in the same way as at the very beginning of the example.

We have a triangular determinant whose elements below the diagonal are equal to zero, and by property 7 it is equal to the product of the diagonal elements. In the end we got ∆ A = 26, therefore the inverse matrix exists.

A11 = 1*(3+1) = 4

A12 = -1*(9+2) = -11

A13 = 1*1 = 1

A21 = -1*(-6) = 6

A22 = 1*(3-0) = 3

A23 = -1*(1+4) = -5

A31 = 1*2 = 2

A32 = -1*(-1) = -1

A33 = 1+(1+6) = 7

3. The next step is to compile a matrix from the resulting additions:

5. Multiply this matrix by the inverse of the determinant, that is, by 1/26:

6. Now we just need to check:

During the test, we received an identity matrix, therefore, the solution was carried out absolutely correctly.

2 way to calculate the inverse matrix.

1. Elementary matrix transformation

2. Inverse matrix through an elementary converter.

Elementary matrix transformation includes:

1. Multiplying a string by a number that is not equal to zero.

2. Adding to any line another line multiplied by a number.

3. Swap the rows of the matrix.

4. Applying a chain of elementary transformations, we obtain another matrix.

A -1 = ?

1. (A|E) ~ (E|A -1 )

2.A -1 * A = E

Let's look at this using a practical example with real numbers.

Exercise: Find the inverse matrix.

Solution:

Let's check:

A little clarification on the solution:

First, we rearranged rows 1 and 2 of the matrix, then multiplied the first row by (-1).

After that, we multiplied the first row by (-2) and added it with the second row of the matrix. Then we multiplied line 2 by 1/4.

The final stage of transformation was multiplying the second line by 2 and adding it with the first. As a result, we have the identity matrix on the left, therefore, the inverse matrix is ​​the matrix on the right.

After checking, we were convinced that the decision was correct.

As you can see, calculating the inverse matrix is ​​very simple.

At the end of this lecture, I would also like to spend a little time on the properties of such a matrix.

This topic is one of the most hated among students. Worse, probably, are the qualifiers.

The trick is that the very concept of an inverse element (and I’m not just talking about matrices) refers us to the operation of multiplication. Even in school curriculum multiplication is considered a complex operation, and matrix multiplication in general separate topic, to which I have a whole paragraph and video tutorial dedicated.

Today we will not go into the details of matrix calculations. Let’s just remember: how matrices are designated, how they are multiplied, and what follows from this.

Review: Matrix Multiplication

First of all, let's agree on notation. A matrix $A$ of size $\left[ m\times n \right]$ is simply a table of numbers with exactly $m$ rows and $n$ columns:

\=\underbrace(\left[ \begin(matrix) ((a)_(11)) & ((a)_(12)) & ... & ((a)_(1n)) \\ (( a)_(21)) & ((a)_(22)) & ... & ((a)_(2n)) \\ ... & ... & ... & ... \\ ((a)_(m1)) & ((a)_(m2)) & ... & ((a)_(mn)) \\\end(matrix) \right])_(n)\]

To avoid accidentally mixing up rows and columns (believe me, in an exam you can confuse a one with a two, let alone some rows), just look at the picture:

Determining indices for matrix cells

What's happening? If you place the standard coordinate system $OXY$ in the upper left corner and direct the axes so that they cover the entire matrix, then each cell of this matrix can be uniquely associated with coordinates $\left(x;y \right)$ - this will be the row number and column number.

Why is the coordinate system placed in the upper left corner? Yes, because it is from there that we begin to read any texts. It's very easy to remember.

Why is the $x$ axis directed downwards and not to the right? Again, it's simple: take a standard coordinate system (the $x$ axis goes to the right, the $y$ axis goes up) and rotate it so that it covers the matrix. This is a 90 degree clockwise rotation - we see the result in the picture.

In general, we have figured out how to determine the indices of matrix elements. Now let's look at multiplication.

Definition. Matrices $A=\left[ m\times n \right]$ and $B=\left[ n\times k \right]$, when the number of columns in the first coincides with the number of rows in the second, are called consistent.

Exactly in that order. One can be confused and say that the matrices $A$ and $B$ form an ordered pair $\left(A;B \right)$: if they are consistent in this order, then it is not at all necessary that $B$ and $A$ those. the pair $\left(B;A \right)$ is also consistent.

Only matched matrices can be multiplied.

Definition. The product of matched matrices $A=\left[ m\times n \right]$ and $B=\left[ n\times k \right]$ is the new matrix $C=\left[ m\times k \right]$ , the elements of which $((c)_(ij))$ are calculated according to the formula:

\[((c)_(ij))=\sum\limits_(k=1)^(n)(((a)_(ik)))\cdot ((b)_(kj))\]

In other words: to get the element $((c)_(ij))$ of the matrix $C=A\cdot B$, you need to take the $i$-row of the first matrix, the $j$-th column of the second matrix, and then multiply in pairs elements from this row and column. Add up the results.

Yes, that’s such a harsh definition. Several facts immediately follow from it:

1. Matrix multiplication, generally speaking, is non-commutative: $A\cdot B\ne B\cdot A$;
2. However, multiplication is associative: $\left(A\cdot B \right)\cdot C=A\cdot \left(B\cdot C \right)$;
3. And even distributively: $\left(A+B \right)\cdot C=A\cdot C+B\cdot C$;
4. And once again distributively: $A\cdot \left(B+C \right)=A\cdot B+A\cdot C$.

The distributivity of multiplication had to be described separately for the left and right sum factor precisely because of the non-commutativity of the multiplication operation.

If it turns out that $A\cdot B=B\cdot A$, such matrices are called commutative.

Among all the matrices that are multiplied by something there, there are special ones - those that, when multiplied by any matrix $A$, again give $A$:

Definition. A matrix $E$ is called identity if $A\cdot E=A$ or $E\cdot A=A$. In the case of a square matrix $A$ we can write:

The identity matrix is ​​a frequent guest in solving matrix equations. And in general, a frequent guest in the world of matrices. :)

And because of this $E$, someone came up with all the nonsense that will be written next.

What is an inverse matrix

Since matrix multiplication is a very labor-intensive operation (you have to multiply a bunch of rows and columns), the concept of an inverse matrix also turns out to be not the most trivial. And requiring some explanation.

Key Definition

Well, it's time to know the truth.

Definition. A matrix $B$ is called the inverse of a matrix $A$ if

The inverse matrix is ​​denoted by $((A)^(-1))$ (not to be confused with the degree!), so the definition can be rewritten as follows:

It would seem that everything is extremely simple and clear. But when analyzing this definition, several questions immediately arise:

1. Does an inverse matrix always exist? And if not always, then how to determine: when it exists and when it does not?
2. And who said that there is exactly one such matrix? What if for some initial matrix $A$ there is a whole crowd of inverses?
3. What do all these “reverses” look like? And how, exactly, should we count them?

As for calculation algorithms, we will talk about this a little later. But we will answer the remaining questions right now. Let us formulate them in the form of separate statements-lemmas.

Basic properties

Let's start with how the matrix $A$ should, in principle, look in order for $((A)^(-1))$ to exist for it. Now we will make sure that both of these matrices must be square, and of the same size: $\left[ n\times n \right]$.

Lemma 1. Given a matrix $A$ and its inverse $((A)^(-1))$. Then both of these matrices are square, and of the same order $n$.

Proof. It's simple. Let the matrix $A=\left[ m\times n \right]$, $((A)^(-1))=\left[ a\times b \right]$. Since the product $A\cdot ((A)^(-1))=E$ exists by definition, the matrices $A$ and $((A)^(-1))$ are consistent in the order shown:

\[\begin(align) & \left[ m\times n \right]\cdot \left[ a\times b \right]=\left[ m\times b \right] \\ & n=a \end( align)\]

This direct consequence from the matrix multiplication algorithm: the coefficients $n$ and $a$ are “transit” and must be equal.

At the same time, the inverse multiplication is also defined: $((A)^(-1))\cdot A=E$, therefore the matrices $((A)^(-1))$ and $A$ are also consistent in the specified order:

\[\begin(align) & \left[ a\times b \right]\cdot \left[ m\times n \right]=\left[ a\times n \right] \\ & b=m \end( align)\]

Thus, without loss of generality, we can assume that $A=\left[ m\times n \right]$, $((A)^(-1))=\left[ n\times m \right]$. However, according to the definition of $A\cdot ((A)^(-1))=((A)^(-1))\cdot A$, therefore the sizes of the matrices strictly coincide:

\[\begin(align) & \left[ m\times n \right]=\left[ n\times m \right] \\ & m=n \end(align)\]

So it turns out that all three matrices - $A$, $((A)^(-1))$ and $E$ - are square matrices of size $\left[ n\times n \right]$. The lemma is proven.

Well, that's already good. We see that only square matrices are invertible. Now let's make sure that the inverse matrix is ​​always the same.

Lemma 2. Given a matrix $A$ and its inverse $((A)^(-1))$. Then this inverse matrix is ​​the only one.

Proof. Let's go by contradiction: let the matrix $A$ have at least two inverses - $B$ and $C$. Then, according to definition, the following equalities are true:

\[\begin(align) & A\cdot B=B\cdot A=E; \\ & A\cdot C=C\cdot A=E. \\ \end(align)\]

From Lemma 1 we conclude that all four matrices - $A$, $B$, $C$ and $E$ - are squares of the same order: $\left[ n\times n \right]$. Therefore, the product is defined:

Since matrix multiplication is associative (but not commutative!), we can write:

\[\begin(align) & B\cdot A\cdot C=\left(B\cdot A \right)\cdot C=E\cdot C=C; \\ & B\cdot A\cdot C=B\cdot \left(A\cdot C \right)=B\cdot E=B; \\ & B\cdot A\cdot C=C=B\Rightarrow B=C. \\ \end(align)\]

We received the only possible variant: two instances of the inverse matrix are equal. The lemma is proven.

The above arguments repeat almost verbatim the proof of the uniqueness of the inverse element for all real numbers $b\ne 0$. The only significant addition is taking into account the dimension of matrices.

However, we still do not know anything about whether every square matrix is ​​invertible. This is where the determinant comes to our aid - this is a key characteristic for everyone square matrices.

Lemma 3. Given a matrix $A$. If its inverse matrix $((A)^(-1))$ exists, then the determinant of the original matrix is ​​nonzero:

\[\left| A\right|\ne 0\]

Proof. We already know that $A$ and $((A)^(-1))$ are square matrices of size $\left[ n\times n \right]$. Therefore, for each of them we can calculate the determinant: $\left| A\right|$ and $\left| ((A)^(-1)) \right|$. However, the determinant of a product is equal to the product of the determinants:

\[\left| A\cdot B \right|=\left| A \right|\cdot \left| B \right|\Rightarrow \left| A\cdot ((A)^(-1)) \right|=\left| A \right|\cdot \left| ((A)^(-1)) \right|\]

But according to the definition, $A\cdot ((A)^(-1))=E$, and the determinant of $E$ is always equal to 1, so

\[\begin(align) & A\cdot ((A)^(-1))=E; \\ & \left| A\cdot ((A)^(-1)) \right|=\left| E\right|; \\ & \left| A \right|\cdot \left| ((A)^(-1)) \right|=1. \\ \end(align)\]

The product of two numbers is equal to one only if each of these numbers is non-zero:

\[\left| A \right|\ne 0;\quad \left| ((A)^(-1)) \right|\ne 0.\]

So it turns out that $\left| A \right|\ne 0$. The lemma is proven.

In fact, this requirement is quite logical. Now we will analyze the algorithm for finding the inverse matrix - and it will become completely clear why, with a zero determinant, no inverse matrix in principle can exist.

But first, let’s formulate an “auxiliary” definition:

Definition. A singular matrix is ​​a square matrix of size $\left[ n\times n \right]$ whose determinant is zero.

Thus, we can claim that every invertible matrix is ​​non-singular.

How to find the inverse of a matrix

Now we will consider a universal algorithm for finding inverse matrices. In general, there are two generally accepted algorithms, and we will also consider the second one today.

The one that will be discussed now is very effective for matrices of size $\left[ 2\times 2 \right]$ and - partially - size $\left[ 3\times 3 \right]$. But starting from the size $\left[ 4\times 4 \right]$ it is better not to use it. Why - now you will understand everything yourself.

Algebraic additions

Get ready. Now there will be pain. No, don’t worry: a beautiful nurse in a skirt, stockings with lace will not come to you and give you an injection in the buttock. Everything is much more prosaic: algebraic additions and Her Majesty the “Union Matrix” come to you.

Let's start with the main thing. Let there be a square matrix of size $A=\left[ n\times n \right]$, whose elements are called $((a)_(ij))$. Then for each such element we can define an algebraic complement:

Definition. Algebraic complement $((A)_(ij))$ to the element $((a)_(ij))$ located in the $i$th row and $j$th column of the matrix $A=\left[ n \times n \right]$ is a construction of the form

\[((A)_(ij))=((\left(-1 \right))^(i+j))\cdot M_(ij)^(*)\]

Where $M_(ij)^(*)$ is the determinant of the matrix obtained from the original $A$ by deleting the same $i$th row and $j$th column.

Again. The algebraic complement to a matrix element with coordinates $\left(i;j \right)$ is denoted as $((A)_(ij))$ and is calculated according to the scheme:

1. First, we delete the $i$-row and $j$-th column from the original matrix. We obtain a new square matrix, and we denote its determinant as $M_(ij)^(*)$.
2. Then we multiply this determinant by $((\left(-1 \right))^(i+j))$ - at first this expression may seem mind-blowing, but in essence we are simply figuring out the sign in front of $M_(ij)^(*) $.
3. We count and get a specific number. Those. the algebraic addition is precisely a number, and not some new matrix, etc.

The matrix $M_(ij)^(*)$ itself is called an additional minor to the element $((a)_(ij))$. And in this sense, the above definition of an algebraic complement is a special case of a more complex definition - what we looked at in the lesson about the determinant.

Important note. Actually, in “adult” mathematics, algebraic additions are defined as follows:

1. We take $k$ rows and $k$ columns in a square matrix. At their intersection we get a matrix of size $\left[ k\times k \right]$ - its determinant is called a minor of order $k$ and is denoted $((M)_(k))$.
2. Then we cross out these “selected” $k$ rows and $k$ columns. Once again you get a square matrix - its determinant is called an additional minor and is denoted $M_(k)^(*)$.
3. Multiply $M_(k)^(*)$ by $((\left(-1 \right))^(t))$, where $t$ is (attention now!) the sum of the numbers of all selected rows and columns . This will be the algebraic addition.

Look at the third step: there is actually a sum of $2k$ terms! Another thing is that for $k=1$ we will get only 2 terms - these will be the same $i+j$ - the “coordinates” of the element $((a)_(ij))$ for which we are looking for an algebraic complement.

So today we're using a slightly simplified definition. But as we will see later, it will be more than enough. The following thing is much more important:

Definition. The allied matrix $S$ to the square matrix $A=\left[ n\times n \right]$ is a new matrix of size $\left[ n\times n \right]$, which is obtained from $A$ by replacing $(( a)_(ij))$ by algebraic additions $((A)_(ij))$:

\\Rightarrow S=\left[ \begin(matrix) ((A)_(11)) & ((A)_(12)) & ... & ((A)_(1n)) \\ (( A)_(21)) & ((A)_(22)) & ... & ((A)_(2n)) \\ ... & ... & ... & ... \\ ((A)_(n1)) & ((A)_(n2)) & ... & ((A)_(nn)) \\\end(matrix) \right]\]

The first thought that arises at the moment of realizing this definition is “how much will have to be counted!” Relax: you will have to count, but not that much. :)

Well, all this is very nice, but why is it necessary? But why.

Main theorem

Let's go back a little. Remember, in Lemma 3 it was stated that the invertible matrix $A$ is always non-singular (that is, its determinant is non-zero: $\left| A \right|\ne 0$).

So, the opposite is also true: if the matrix $A$ is not singular, then it is always invertible. And there is even a search scheme for $((A)^(-1))$. Check it out:

Inverse matrix theorem. Let a square matrix $A=\left[ n\times n \right]$ be given, and its determinant is nonzero: $\left| A \right|\ne 0$. Then the inverse matrix $((A)^(-1))$ exists and is calculated by the formula:

\[((A)^(-1))=\frac(1)(\left| A \right|)\cdot ((S)^(T))\]

And now - everything is the same, but in legible handwriting. To find the inverse matrix, you need:

1. Calculate the determinant $\left| A \right|$ and make sure it is non-zero.
2. Construct the union matrix $S$, i.e. count 100500 algebraic additions $((A)_(ij))$ and place them in place $((a)_(ij))$.
3. Transpose this matrix $S$, and then multiply it by some number $q=(1)/(\left| A \right|)\;$.

That's all! The inverse matrix $((A)^(-1))$ has been found. Let's look at examples:

\[\left[ \begin(matrix) 3 & 1 \\ 5 & 2 \\\end(matrix) \right]\]

Solution. Let's check the reversibility. Let's calculate the determinant:

\[\left| A\right|=\left| \begin(matrix) 3 & 1 \\ 5 & 2 \\\end(matrix) \right|=3\cdot 2-1\cdot 5=6-5=1\]

The determinant is different from zero. This means the matrix is ​​invertible. Let's create a union matrix:

Let's calculate the algebraic additions:

\[\begin(align) & ((A)_(11))=((\left(-1 \right))^(1+1))\cdot \left| 2 \right|=2; \\ & ((A)_(12))=((\left(-1 \right))^(1+2))\cdot \left| 5 \right|=-5; \\ & ((A)_(21))=((\left(-1 \right))^(2+1))\cdot \left| 1 \right|=-1; \\ & ((A)_(22))=((\left(-1 \right))^(2+2))\cdot \left| 3\right|=3. \\ \end(align)\]

Please note: the determinants |2|, |5|, |1| and |3| are determinants of matrices of size $\left[ 1\times 1 \right]$, and not modules. Those. If there were negative numbers in the determinants, there is no need to remove the “minus”.

In total, our union matrix looks like this:

\[((A)^(-1))=\frac(1)(\left| A \right|)\cdot ((S)^(T))=\frac(1)(1)\cdot ( (\left[ \begin(array)(*(35)(r)) 2 & -5 \\ -1 & 3 \\\end(array) \right])^(T))=\left[ \begin (array)(*(35)(r)) 2 & -1 \\ -5 & 3 \\\end(array) \right]\]

OK it's all over Now. The problem is solved.

Answer. $\left[ \begin(array)(*(35)(r)) 2 & -1 \\ -5 & 3 \\\end(array) \right]$

Task. Find the inverse matrix:

\[\left[ \begin(array)(*(35)(r)) 1 & -1 & 2 \\ 0 & 2 & -1 \\ 1 & 0 & 1 \\\end(array) \right] \]

Solution. We calculate the determinant again:

\[\begin(align) & \left| \begin(array)(*(35)(r)) 1 & -1 & 2 \\ 0 & 2 & -1 \\ 1 & 0 & 1 \\\end(array) \right|=\begin(matrix ) \left(1\cdot 2\cdot 1+\left(-1 \right)\cdot \left(-1 \right)\cdot 1+2\cdot 0\cdot 0 \right)- \\ -\left (2\cdot 2\cdot 1+\left(-1 \right)\cdot 0\cdot 1+1\cdot \left(-1 \right)\cdot 0 \right) \\\end(matrix)= \ \ & =\left(2+1+0 \right)-\left(4+0+0 \right)=-1\ne 0. \\ \end(align)\]

The determinant is nonzero—the matrix is ​​invertible. But now it’s going to be really tough: we need to count as many as 9 (nine, motherfucker!) algebraic additions. And each of them will contain the determinant $\left[ 2\times 2 \right]$. Flew:

\[\begin(matrix) ((A)_(11))=((\left(-1 \right))^(1+1))\cdot \left| \begin(matrix) 2 & -1 \\ 0 & 1 \\\end(matrix) \right|=2; \\ ((A)_(12))=((\left(-1 \right))^(1+2))\cdot \left| \begin(matrix) 0 & -1 \\ 1 & 1 \\\end(matrix) \right|=-1; \\ ((A)_(13))=((\left(-1 \right))^(1+3))\cdot \left| \begin(matrix) 0 & 2 \\ 1 & 0 \\\end(matrix) \right|=-2; \\ ... \\ ((A)_(33))=((\left(-1 \right))^(3+3))\cdot \left| \begin(matrix) 1 & -1 \\ 0 & 2 \\\end(matrix) \right|=2; \\ \end(matrix)\]

In short, the union matrix will look like this:

Therefore, the inverse matrix will be:

\[((A)^(-1))=\frac(1)(-1)\cdot \left[ \begin(matrix) 2 & -1 & -2 \\ 1 & -1 & -1 \\ -3 & 1 & 2 \\\end(matrix) \right]=\left[ \begin(array)(*(35)(r))-2 & -1 & 3 \\ 1 & 1 & -1 \ \2 & 1 & -2 \\\end(array) \right]\]

That's it. Here is the answer.

Answer. $\left[ \begin(array)(*(35)(r)) -2 & -1 & 3 \\ 1 & 1 & -1 \\ 2 & 1 & -2 \\\end(array) \right ]$

As you can see, at the end of each example we performed a check. In this regard, an important note:

Don't be lazy to check. Multiply the original matrix by the found inverse matrix - you should get $E$.

Performing this check is much easier and faster than looking for an error in further calculations when, for example, you are solving a matrix equation.

Alternative way

As I said, the inverse matrix theorem works great for sizes $\left[ 2\times 2 \right]$ and $\left[ 3\times 3 \right]$ (in the latter case, it’s not so “great” "), but for matrices large sizes the sadness begins.

But don’t worry: there is an alternative algorithm with which you can calmly find the inverse even for the matrix $\left[ 10\times 10 \right]$. But, as often happens, to consider this algorithm we need a little theoretical background.

Elementary transformations

Among all possible matrix transformations, there are several special ones - they are called elementary. There are exactly three such transformations:

1. Multiplication. You can take the $i$th row (column) and multiply it by any number $k\ne 0$;
2. Addition. Add to the $i$-th row (column) any other $j$-th row (column), multiplied by any number $k\ne 0$ (you can, of course, do $k=0$, but what's the point? ? Nothing will change).
3. Rearrangement. Take the $i$th and $j$th rows (columns) and swap places.

Why these transformations are called elementary (for large matrices they do not look so elementary) and why there are only three of them - these questions are beyond the scope of today's lesson. Therefore, we will not go into details.

Another thing is important: we have to perform all these perversions on the adjoint matrix. Yes, yes: you heard right. Now there will be one more definition - the last one in today's lesson.

Adjoint matrix

Surely at school you solved systems of equations using the addition method. Well, there, subtract another from one line, multiply some line by a number - that’s all.

So: now everything will be the same, but in an “adult” way. Ready?

Definition. Let a matrix $A=\left[ n\times n \right]$ and an identity matrix $E$ of the same size $n$ be given. Then the adjoint matrix $\left[ A\left| E\right. \right]$ is a new matrix of size $\left[ n\times 2n \right]$ that looks like this:

\[\left[ A\left| E\right. \right]=\left[ \begin(array)(rrrr|rrrr)((a)_(11)) & ((a)_(12)) & ... & ((a)_(1n)) & 1 & 0 & ... & 0 \\((a)_(21)) & ((a)_(22)) & ... & ((a)_(2n)) & 0 & 1 & ... & 0 \\... & ... & ... & ... & ... & ... & ... & ... \\((a)_(n1)) & ((a)_(n2)) & ... & ((a)_(nn)) & 0 & 0 & ... & 1 \\\end(array) \right]\]

In short, we take the matrix $A$, and on the right we assign to it the identity matrix $E$ the right size, we separate them with a vertical line for beauty - here you have the attached one. :)

What's the catch? Here's what:

Theorem. Let the matrix $A$ be invertible. Consider the adjoint matrix $\left[ A\left| E\right. \right]$. If using elementary string conversions bring it to the form $\left[ E\left| B\right. \right]$, i.e. by multiplying, subtracting and rearranging rows to obtain from $A$ the matrix $E$ on the right, then the matrix $B$ obtained on the left is the inverse of $A$:

\[\left[ A\left| E\right. \right]\to \left[ E\left| B\right. \right]\Rightarrow B=((A)^(-1))\]

It's that simple! In short, the algorithm for finding the inverse matrix looks like this:

1. Write the adjoint matrix $\left[ A\left| E\right. \right]$;
2. Perform elementary string conversions until $E$ appears instead of $A$;
3. Of course, something will also appear on the left - a certain matrix $B$. This will be the opposite;
4. PROFIT!:)

Of course, this is much easier said than done. So let's look at a couple of examples: for sizes $\left[ 3\times 3 \right]$ and $\left[ 4\times 4 \right]$.

Task. Find the inverse matrix:

\[\left[ \begin(array)(*(35)(r)) 1 & 5 & 1 \\ 3 & 2 & 1 \\ 6 & -2 & 1 \\\end(array) \right]\ ]

Solution. We create the adjoint matrix:

\[\left[ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 3 & 2 & 1 & 0 & 1 & 0 \\ 6 & -2 & 1 & 0 & 0 & 1 \\\end(array) \right]\]

Since the last column of the original matrix is ​​filled with ones, subtract the first row from the rest:

\[\begin(align) & \left[ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 3 & 2 & 1 & 0 & 1 & 0 \\ 6 & - 2 & 1 & 0 & 0 & 1 \\\end(array) \right]\begin(matrix) \downarrow \\ -1 \\ -1 \\\end(matrix)\to \\ & \to \left [ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 2 & -3 & 0 & -1 & 1 & 0 \\ 5 & -7 & 0 & -1 & 0 & 1 \\\end(array) \right] \\ \end(align)\]

There are no more units, except for the first line. But we don’t touch it, otherwise the newly removed units will begin to “multiply” in the third column.

But we can subtract the second line twice from the last - we get one in the lower left corner:

\[\begin(align) & \left[ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 2 & -3 & 0 & -1 & 1 & 0 \\ 5 & -7 & 0 & -1 & 0 & 1 \\\end(array) \right]\begin(matrix) \ \\ \downarrow \\ -2 \\\end(matrix)\to \\ & \left [ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 2 & -3 & 0 & -1 & 1 & 0 \\ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right] \\ \end(align)\]

Now we can subtract the last row from the first and twice from the second - this way we “zero” the first column:

\[\begin(align) & \left[ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 2 & -3 & 0 & -1 & 1 & 0 \\ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right]\begin(matrix) -1 \\ -2 \\ \uparrow \\\end(matrix)\to \\ & \ to \left[ \begin(array)(rrr|rrr) 0 & 6 & 1 & 0 & 2 & -1 \\ 0 & -1 & 0 & -3 & 5 & -2 \\ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right] \\ \end(align)\]

Multiply the second line by −1, and then subtract it 6 times from the first and add 1 time to the last:

\[\begin(align) & \left[ \begin(array)(rrr|rrr) 0 & 6 & 1 & 0 & 2 & -1 \\ 0 & -1 & 0 & -3 & 5 & -2 \ \ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right]\begin(matrix) \ \\ \left| \cdot \left(-1 \right) \right. \\ \ \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrr|rrr) 0 & 6 & 1 & 0 & 2 & -1 \\ 0 & 1 & 0 & 3 & -5 & 2 \\ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right]\begin(matrix) -6 \\ \updownarrow \\ +1 \\\end (matrix)\to \\ & \to \left[ \begin(array)(rrr|rrr) 0 & 0 & 1 & -18 & 32 & -13 \\ 0 & 1 & 0 & 3 & -5 & 2 \\ 1 & 0 & 0 & 4 & -7 & 3 \\\end(array) \right] \\ \end(align)\]

All that remains is to swap lines 1 and 3:

\[\left[ \begin(array)(rrr|rrr) 1 & 0 & 0 & 4 & -7 & 3 \\ 0 & 1 & 0 & 3 & -5 & 2 \\ 0 & 0 & 1 & - 18 & 32 & -13 \\\end(array) \right]\]

Ready! On the right is the required inverse matrix.

Answer. $\left[ \begin(array)(*(35)(r))4 & -7 & 3 \\ 3 & -5 & 2 \\ -18 & 32 & -13 \\\end(array) \right ]$

Task. Find the inverse matrix:

\[\left[ \begin(matrix) 1 & 4 & 2 & 3 \\ 1 & -2 & 1 & -2 \\ 1 & -1 & 1 & 1 \\ 0 & -10 & -2 & -5 \\\end(matrix) \right]\]

Solution. We compose the adjoint again:

\[\left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 1 & -2 & 1 & -2 & 0 & 1 & 0 & 0 \ \ 1 & -1 & 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & -10 & -2 & -5 & 0 & 0 & 0 & 1 \\\end(array) \right]\]

Let's cry a little, be sad about how much we have to count now... and start counting. First, let’s “zero out” the first column by subtracting row 1 from rows 2 and 3:

\[\begin(align) & \left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 1 & -2 & 1 & -2 & 0 & 1 & 0 & 0 \\ 1 & -1 & 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & -10 & -2 & -5 & 0 & 0 & 0 & 1 \\\end(array) \right]\begin(matrix) \downarrow \\ -1 \\ -1 \\ \ \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 0 & -6 & -1 & -5 & -1 & 1 & 0 & 0 \\ 0 & -5 & -1 & -2 & -1 & 0 & 1 & 0 \\ 0 & -10 & -2 & -5 & 0 & 0 & 0 & 1 \\\end(array) \right] \\ \end(align)\]

We see too many “cons” in lines 2-4. Multiply all three rows by −1, and then burn out the third column by subtracting row 3 from the rest:

\[\begin(align) & \left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 0 & -6 & -1 & -5 & - 1 & 1 & 0 & 0 \\ 0 & -5 & -1 & -2 & -1 & 0 & 1 & 0 \\ 0 & -10 & -2 & -5 & 0 & 0 & 0 & 1 \\ \end(array) \right]\begin(matrix) \ \\ \left| \cdot \left(-1 \right) \right. \\ \left| \cdot \left(-1 \right) \right. \\ \left| \cdot \left(-1 \right) \right. \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 0 & 6 & 1 & 5 & ​​1 & -1 & 0 & 0 \\ 0 & 5 & 1 & 2 & 1 & 0 & -1 & 0 \\ 0 & 10 & 2 & 5 & 0 & 0 & 0 & -1 \\\end (array) \right]\begin(matrix) -2 \\ -1 \\ \updownarrow \\ -2 \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr| rrrr) 1 & -6 & 0 & -1 & -1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 3 & 0 & -1 & 1 & 0 \\ 0 & 5 & 1 & 2 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end(array) \right] \\ \end(align)\]

Now is the time to “fry” the last column of the original matrix: subtract line 4 from the rest:

\[\begin(align) & \left[ \begin(array)(rrrr|rrrr) 1 & -6 & 0 & -1 & -1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 3 & 0 & -1 & 1 & 0 \\ 0 & 5 & 1 & 2 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end(array ) \right]\begin(matrix) +1 \\ -3 \\ -2 \\ \uparrow \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr|rrrr) 1 & -6 & 0 & 0 & -3 & 0 & 4 & -1 \\ 0 & 1 & 0 & 0 & 6 & -1 & -5 & 3 \\ 0 & 5 & 1 & 0 & 5 & 0 & -5 & 2 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end(array) \right] \\ \end(align)\]

Final throw: “burn out” the second column by subtracting line 2 from lines 1 and 3:

\[\begin(align) & \left[ \begin(array)(rrrr|rrrr) 1 & -6 & 0 & 0 & -3 & 0 & 4 & -1 \\ 0 & 1 & 0 & 0 & 6 & -1 & -5 & 3 \\ 0 & 5 & 1 & 0 & 5 & 0 & -5 & 2 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end( array) \right]\begin(matrix) 6 \\ \updownarrow \\ -5 \\ \ \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr|rrrr) 1 & 0 & 0 & 0 & 33 & -6 & -26 & -17 \\ 0 & 1 & 0 & 0 & 6 & -1 & -5 & 3 \\ 0 & 0 & 1 & 0 & -25 & 5 & 20 & -13 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end(array) \right] \\ \end(align)\]

And again the identity matrix is ​​on the left, which means the inverse is on the right. :)

Answer. $\left[ \begin(matrix) 33 & -6 & -26 & 17 \\ 6 & -1 & -5 & 3 \\ -25 & 5 & 20 & -13 \\ -2 & 0 & 2 & - 1 \\\end(matrix) \right]$

Matrix algebra - Inverse matrix

inverse matrix

Inverse matrix is called a matrix which, when multiplied both on the right and on the left by this matrix gives the identity matrix.
Let us denote the inverse matrix of the matrix A through , then according to definition we get:

Where E– identity matrix.
Square matrix called not special (non-degenerate) if its determinant is not zero. Otherwise it is called special (degenerate) or singular.

The theorem holds: Every non-singular matrix has an inverse matrix.

The operation of finding the inverse matrix is ​​called appeal matrices. Let's consider the matrix inversion algorithm. Let a non-singular matrix be given n-th order:

where Δ = det A ≠ 0.

Algebraic addition of an element matrices n-th order A is called the determinant of a matrix taken with a certain sign ( n–1)th order obtained by deleting i-th line and j th matrix column A:

Let's create the so-called attached matrix:

where are the algebraic complements of the corresponding elements of the matrix A.
Note that algebraic additions of matrix row elements A are placed in the corresponding columns of the matrix à , that is, the matrix is ​​transposed at the same time.
By dividing all the elements of the matrix à by Δ – the value of the matrix determinant A, we get the inverse matrix as a result:

Let's note the row special properties inverse matrix:
1) for a given matrix A its inverse matrix is the only one;
2) if there is an inverse matrix, then right reverse And left reverse the matrices coincide with it;
3) a singular (singular) square matrix does not have an inverse matrix.

Basic properties of an inverse matrix:
1) the determinant of the inverse matrix and the determinant of the original matrix are reciprocals;
2) the inverse matrix of the product of square matrices is equal to the product of the inverse matrix of factors, taken in reverse order:

3) the transposed inverse matrix is ​​equal to the inverse matrix of the given transposed matrix:

EXAMPLE Calculate the inverse of the given matrix.